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EMI testing with 10 V/M immunity

Discussion in 'Electronic Design' started by [email protected], Mar 23, 2007.

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  1. Guest

    Hi Group,

    I bit of confusion reviewing the E(V/M) equation for power gives:

    P(watts) = (E^2*r^2)/(30xantenna gain), at 10V per meter this is 14.64
    Watts of input power.

    The equation, out of the EMC textbook, does not include frequency.

    If I put 14.64 watts into a dipole at 100 MHz and measure the V/M at
    three meters and then repeated this with the frequency moved to 300
    MHz then both would read the same V/M?

    I thought that the higher the frequency, more of the energy is lost so
    an equal input power at 100 MHz will have less space loss than an
    equal input at 300 MHz. This will result in a higher level of
    interference to the device under test at 3 meters.

    pdrunen
     
  2. Fred Bloggs

    Fred Bloggs Guest

    The relationship between source power and field strength for a wave
    propagating in free space is derived from the Poynting vector to give
    S=|E|^2/Zo W/m^2 where Zo is the impedance of free space ~377 ohms.
    Power density at a distance R, due to source power Po, is also
    |S|=Po*G/(4pi*R^2) where G is the antenna gain, so that the E field
    strength is computed from |E|=sqrt(|S|*Zo). This makes
    Po=(4pi*R^2)/G*|S|=((4pi*R^2)/G)*|E|^2/Zo.
    Your thinking is not right. There are two propagation loss factors: 1)
    spherical spreading and 2) atmospheric absorption, only spherical
    spreading is significant at short distances and you already have the
    equation for that. You are missing out on a fundamental relationship
    between antenna peak gain,G, and effective aperture area, A, which due
    to arguments relating to reciprocity of xmit and receive can be shown to
    have a constant ratio of G/A=4pi/lamda^2 where lamda is the wavelength.
    Since you are effectively maintaining constant A while increasing the
    frequency by a factor of x3, the peak gain at 300MHz will be 9x that at
    100MHz. Therefore your |S| will increase by x9 due to G, and your |E|
    will increase by sqrt(|S|) or by x3, for the the same Po.
     
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